integral of (19)/((1-x^2)^{3/2)}

解答

\int \frac{19}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{19 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}} + C

求解步骤

\int \frac{19}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 19 \cdot \int \frac{1}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

使用三角换元法

= 19 \cdot \int sec^{2} (u) d u

使用积分常见定则:

\int sec^{2} (u) d u = tan (u)

= 19 tan (u)

u = arcsin (x)

代回

= 19 tan (arcsin (x))

化简

19 tan (arcsin (x)) : \frac{19 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}}

= \frac{19 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}}

解答补常数

= \frac{19 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}} + C

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